Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $p \neq 0$. $r = \dfrac{p + 1}{p - 3} \times \dfrac{p^2 - 3p}{p^2 - 2p - 3} $
Answer: First factor the quadratic. $r = \dfrac{p + 1}{p - 3} \times \dfrac{p^2 - 3p}{(p + 1)(p - 3)} $ Then factor out any other terms. $r = \dfrac{p + 1}{p - 3} \times \dfrac{p(p - 3)}{(p + 1)(p - 3)} $ Then multiply the two numerators and multiply the two denominators. $r = \dfrac{ (p + 1) \times p(p - 3) } { (p - 3) \times (p + 1)(p - 3) } $ $r = \dfrac{ p(p + 1)(p - 3)}{ (p - 3)(p + 1)(p - 3)} $ Notice that $(p - 3)$ and $(p + 1)$ appear in both the numerator and denominator so we can cancel them. $r = \dfrac{ p\cancel{(p + 1)}(p - 3)}{ (p - 3)\cancel{(p + 1)}(p - 3)} $ We are dividing by $p + 1$ , so $p + 1 \neq 0$ Therefore, $p \neq -1$ $r = \dfrac{ p\cancel{(p + 1)}\cancel{(p - 3)}}{ \cancel{(p - 3)}\cancel{(p + 1)}(p - 3)} $ We are dividing by $p - 3$ , so $p - 3 \neq 0$ Therefore, $p \neq 3$ $r = \dfrac{p}{p - 3} ; \space p \neq -1 ; \space p \neq 3 $